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In calculus, a critical point is where the first derivative of a function is zero or undefined. Finding these points is crucial because they often indicate where the function is changing direction, which can help us identify peaks and valleys in the graph. For the function \( g(t) = -3t^2 + 9t + 5 \), we start by calculating the first derivative, which is \( g'(t) = -6t + 9 \). To find the critical points, we set this equation to zero: \( -6t + 9 = 0 \). Solving for \( t \), we get \( t = \frac{3}{2} \). This point is critical because it is where the slope of the tangent to the graph is zero, indicating a potential local extreme.

A function is increasing on an interval if the first derivative is positive over that interval, and decreasing if the derivative is negative. To determine where \( g(t) = -3t^2 + 9t + 5 \) is increasing or decreasing, we consider intervals around the critical point \( t = \frac{3}{2} \). We test points in the intervals \((-\infty, \frac{3}{2})\) and \((\frac{3}{2}, \infty)\).

• Substituting \( t = 0 \) in the first derivative \( g'(t) = -6t + 9 \) gives \( g'(0) = 9 \), which is positive, indicating that the function is increasing on \((-\infty, \frac{3}{2})\).
• Substituting \( t = 2 \) gives \( g'(2) = -3 \), which is negative, indicating the function is decreasing on \((\frac{3}{2}, \infty)\).

These results show the function behaves differently on each side of the critical point.

Local extrema of a function are the highest or lowest values it takes in a nearby region. They occur at critical points where the function changes direction. For \( g(t) = -3t^2 + 9t + 5 \), we determined earlier that \( t = \frac{3}{2} \) is a critical point. Evaluating the function at this point, \( g(\frac{3}{2}) = 11.25 \), we find a local maximum because the function changes from increasing to decreasing.
Since \( g(t) \) tends to \(-\infty\) as \( t \) approaches \(\pm\infty\), there is no absolute minimum; the function continuously decreases. However, the local maximum at \( t = \frac{3}{2} \) is also the absolute maximum, so \( g(\frac{3}{2}) = 11.25 \) is the highest value that \( g(t) \) reaches for all real values of \( t \). This analysis helps in understanding the overall shape of the function's graph.